Multiplying a Fourier series by cos(2x) and integrating over one full period leaves only the a2 term, because every other sine and cosine in the series integrates to zero against cos(2x).

A periodic signal is 5 on (0, π) and 1 on (π, 2π), repeating every 2π. What is the constant term a0/2 of its Fourier series?

A square wave is -1 on (-pi, 0) and +1 on (0, pi), repeating every 2 pi. It is odd. What does that say about its cosine coefficients?

• Every an is 0
• Every bn is 0
• a0 is infinite

The sawtooth f(x) = x on (-pi, pi) has Fourier sine coefficients bn = 2(-1)^(n+1)/n. What is b4?

What is a partial sum of a Fourier series?

• The series stopped after a few terms
• The sum of all the jump sizes in the wave
• The constant average term a0 over 2 alone

Near a jump discontinuity in the original function, what do the Fourier partial sums do?

• Converge smoothly to the jump once enough terms are kept
• Overshoot by about 9% of the jump, however many terms
• Land exactly on the function's value at the jump
• Drop to zero right at the jump

A square wave jumps from -1 to +1 at x = 0. What value does its Fourier series take at x = 0?

• 1
• 0
• -1
• It does not converge there

If you multiply a periodic function's Fourier series by sin(3x) and integrate over one full period, which coefficient does that isolate?

• a0
• a3
• b3
• b1

Near the square wave jump, the partial sums overshoot no matter how many terms you keep. What should you conclude?

• The coefficients must have been computed wrongly
• Overshoot near jumps is built into the method
• Only even functions ever overshoot at jumps

A circuit is driven by a square wave. How does the series help solve it?

• It replaces the whole circuit with a single resistor
• It splits the drive into tones solved one at a time
• It removes the need for any differential equation